Given a right triangle with lengths that are reciprocals of integers, what is the smallest possible sum of these the integers?

In other words, given a right triangle with lengths 1/a, 1/b, and 1/c, where a, b, and c are all integers, what is the lowest value of a+b+c? Also, prove it.

Taken from CAML, which did not ask for a proof.

The idea in solving this one is to multiply each side by a²b²c² and factor and divide by c² to get a² + b² = (ab/c)²

Since the left side is a perfect square, the right side must be a perfect square. So a and b must be the lesser two from a Pythagorean triple.

The smallest pythagorean triple where all of the values are over 0 but integers is 3, 4, 5.

Putting x3 and x4 (allowing for mutliples of this sequence) in for a and b gives c²(x²5²)=x²3²x²4²

Dividing out x² gives (5*c)² = (3*4*x)²

The least possible value for x is 5 (if c is 12), so a=3(5) b=4(5) and c is 12. So 1/(15²) + 1/(20²) = 1/(12²), and so a = 15, b = 20, c = 12 is the lowest solution, and 15 + 20 + 12 = 47 is the lowest value for a + b + c

Posted by Gamer on 2003-08-05 13:22:11